Question

If tangent to the curve $x=a{t}^2,y=2at$ is perpendicular to x-axis then its point of contact is

• A. $\left(a,a\right)$
• B. $\left(0,a\right)$
• C. $\left(0,0\right)$
• D. $\left(a,0\right)$

Answer 1

The correct option is C

$\left(0,0\right)$

Explanation for the correct option

Given that the curve $x=a{t}^2,y=2at$ is perpendicular to x-axis

$\Rightarrow$slope of the tangent $\frac{dy}{dx}=\infty$

Applying differentiation on both sides of the equation $x=a{t}^2$ we get,

$\begin{array}{crcll}& \frac{dx}{dt}& =& \frac{d}{dt}\left(a{t}^2\right)& \\ & & =& a\frac{d}{dt}\left(t^2\right)& \\ \Rightarrow & \frac{dx}{dt}& =& 2at& ...\left[1\right]\end{array}$

Applying differentiation on both sides of the equation $y=2at$ we get,

$\begin{array}{cccll}& \frac{dy}{dt}& =& \frac{d}{dt}\left(2at\right)& \\ & & =& 2a\frac{d}{dt}\left(t\right)& \\ \Rightarrow & \frac{dy}{dt}& =& 2a& ...\left[2\right]\end{array}$

$\begin{array}{ccccc}& \frac{dy}{dx}& =& \frac{dy}{dt}\times \frac{dt}{dx}& ...\left[3\right]\end{array}$

Substituting $\left[1\right]$ and $\left[2\right]$ in $\left[3\right]$, we get,

$\begin{array}{ccl}\frac{dy}{dx}& =& 2a\times \frac{1}{2at}\\ & =& \frac{1}{t}\end{array}$

We know that $\frac{dy}{dx}=\infty$$\Rightarrow \frac{1}{t}=\infty$

$\Rightarrow t=\frac{1}{\infty }=0$

$\begin{array}{crcl}\therefore & x& =& a{t}^2\\ & & =& a\left(0\right)\\ \Rightarrow & x& =& 0\\ & y& =& 2at\\ & & =& 2a\left(0\right)\\ \Rightarrow & y& =& 0\end{array}$

Therefore, the point of contact is $\left(x,y\right)=\left(0,0\right)$

Hence the correct option is option(C) i.e. $\left(0,0\right)$